(1)已知sinx+cosx=根号2/2,求sin^4x+cos^4x
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/05/14 14:58:19
(1)已知sinx+cosx=根号2/2,求sin^4x+cos^4x
(2)某x在第三象限,化简根号(1+tanx)^2+(1-tanx)^2
(3)化简(2cosθ/根号1-sin^2θ)+根号1-cos^2θ/sinθ
(2)某x在第三象限,化简根号(1+tanx)^2+(1-tanx)^2
(3)化简(2cosθ/根号1-sin^2θ)+根号1-cos^2θ/sinθ
sinx+cosx=√2/2
(sinx+cosx)^2=1/2
(sinx)^2+2sinxcosx+(cosx)^2=1/2
1+2sinxcosx=1/2
2sinxcosx=-1/2
sinxcosx=-1/4
sin^4x+cos^4x
=sin^4x+2(sinxcosx)^2+cos^4x -2(sinxcosx)^2
=[sin^2x+cos^2x ]^2-2(sinxcosx)^2
=1^2-2*(1/4)^2
=1-1/8
=7/8
√[(1+tanx)^2+(1-tanx)^2]
=√[1+tan^2x+2tanx+1+tan^2x-2tanx]
=√[2+2tan^2x]
=√[2(1+tan^2x)]
=√[2sec^2x)]
=-secx√2
2cosθ/√(1-sin^2θ)+√(1-cos^2θ)/sinθ
=2√[cos^2θ/(1-sin^2θ]+√[(1-cos^2θ)/sin^2θ]
=2√[cos^2θ/cos^2θ]+√[sin^2θ/sin^2θ]
=2+1
=3
(sinx+cosx)^2=1/2
(sinx)^2+2sinxcosx+(cosx)^2=1/2
1+2sinxcosx=1/2
2sinxcosx=-1/2
sinxcosx=-1/4
sin^4x+cos^4x
=sin^4x+2(sinxcosx)^2+cos^4x -2(sinxcosx)^2
=[sin^2x+cos^2x ]^2-2(sinxcosx)^2
=1^2-2*(1/4)^2
=1-1/8
=7/8
√[(1+tanx)^2+(1-tanx)^2]
=√[1+tan^2x+2tanx+1+tan^2x-2tanx]
=√[2+2tan^2x]
=√[2(1+tan^2x)]
=√[2sec^2x)]
=-secx√2
2cosθ/√(1-sin^2θ)+√(1-cos^2θ)/sinθ
=2√[cos^2θ/(1-sin^2θ]+√[(1-cos^2θ)/sin^2θ]
=2√[cos^2θ/cos^2θ]+√[sin^2θ/sin^2θ]
=2+1
=3
(1)已知sinx+cosx=根号2/2,求sin^4x+cos^4x
已知sinx-cosx=1/2,求sin^4x+cos^4x的值
已知sinx+cosx=根号2求下列各式的值 sin^4x-cos^4x
已知(3sinx-2cosx)/(sinx+cosx)=4/3,求2/3sin²x-1/4cos²x
已知sinx+cosx/sinx-cosx=2,求2cos(π-x)-3sin(π+x)/4cos(-x)+sin(2π
已知3sin x/2-cosx/2等于0 (1)求tanx的值 (2)求cos2x/根号2cos(派/4+x)sinx的
已知sinx+cosx=2分之根号2求(1/sin^2x)+(1/cos^2x)的值
-4sinx+2cosx=1,sin平方x+cos平方x=1,求sinx,cosx
已知sin(3π-x)-cos(5π+x)=1/2-(根号下3)/2,其中x属于(0,π),求sinx,cosx,x的值
已知(1+tanx)/(1-tanx) =3+2根号2 ,求(sin^2 x+根号2sinx*cosx-cos^2 x)
已知2sin²x-cos²x+sinxcosx-4sinx+2cosx=0求(sin2x+1)/(1
sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x